1. Interrupting fermentation with a one minute
pasteurization at 140 F at the desired alcohol level;
2. Brewing a low-gravity beer and using yeast of the strain
Saccharomyces iudwiggii, which will ferment only simple sugars;
3. Mixing the yeast and wort at 32 F and filtering the yeast
out after a rest period (duration not specified).
The three physical methods noted were:
1. Heating the finished beer at atmospheric pressure. The
authors noted damage to the beer that made this method
undesirable;
2. Reverse osmosis. Removing alcohol by passage through
diaphragms by use of pressure gradients;
3. Dialysis. Passing the beer through fibers that are
bathed in a counterflow dialysate that produces a concentration
gradient.
The recipe I used was for a pale ale-type beer, and consisted of the following:
3# Domestic Malted Barley
1.75# Munich Malt
1.25# Cara-pils
12 ozs. Canadian Wheat
6 ozs. English Crystal (80L)
2 ozs. Domestic Crystal (120L)
.88 ozs. Northern Brewers (8.2% A)--60 mins.
2.69 ozs. Liberty (2.7% A) --10 mins.
Wyeast American Ale yeast (#1056)
The mash was a standard infusion mash, except that the
Cara-pils was not added until the mash was brought to 158 F.
The boil was for 90 mins. A 300 ml yeast starter and 3 tsp.
yeast food were pitched into 6 gals. The O.G. was 1.031.
Primary fermentation lasted for four days at 68 F. Secondary fermentation was for another ten days at 68 F. F.G. was 1.012.
270.4 fl. ozs. was separated from the main batch after fermentation was complete for use in the "no alcohol experiment".
By inverting the bottles, the fluid that did not freeze (including the alcohol) was at the cap-end of the bottle, making it easier to pour it off. The biggest surprise of this whole process was that when I opened the cap on the bottles, they virtually exploded. After the first bottle, and cleaning the sink, cabinets, walls and ceilings of alcohol sludge, I opened the remaining three bottles underneath a plastic bucket. Each one of them exploded. I estimate that 770 ml of the alcohol sludge (I will call the "extract") was lost on the walls and down the sink. I did manage to save 950 ml of the extract.
There was an unexpected problem when I removed the 950 ml extract. I had been forewarned that this freezing method removed a lot of body and hop bitterness (thus the 1# Cara-pils and 35 IBUs), but no one had mentioned the loss of color. This was supposed to be a pale ale, and had been the appropriate color when I froze the beer. But when the extract was removed the beer remaining frozen in the 2 liter bottles was almost as clear as ice--no color at all. The extract, on the other hand, was as dark as a stout. The extract also had a S.G. of 1.039 and smelled like alcohol and HOPS (capitalized on purpose and for effect).
So, I was left with the problem of a colorless, no alcohol beer. This is when I decided to employ the third method of de-alcoholization--heat. Numerous people had warned me about the bad effects of heat on the beer for alcohol removal, and so I decided to heat only the extract. After all, that's where the alcohol was. After I removed the alcohol from the extract, I planned to put the extract back in the beer to give it color again.
Those suggesting heat as an alcohol removal method instructed that alcohol evaporates at 78-82 C (I did not check this fact). Therefore, I heated the 950 ml of extract at 174-178 F for 13 minutes, which resulted in 600 ml of extract at 1.069. After cooling, I returned this extract to the melting beer.
O.G. 1.031 F.G. 1.012
A%(by weight)=76.08(OG-FG)/(1.775-OG)
=76.08(1.031-1.012)/(1.775-1.031)
=76.08(.019)/1.744
=1.996574585
A%(by volume)=A%(weight)[FG/.794]
=1.996574585[1.012/.794]
=2.544752491
Thus the beer had a very low alcohol content of
approximately 2.5% even before doing anything to it.
I next computed the volume of actual alcohol in the beer. The total beer volume was 244 ozs. (7,222.4 ml), and so I multiplied that number by the percent alcohol by volume to determine the number of milliliters of alcohol in the beer:
7,222.4 ml beer x .02544752491 = 203.6779104 ml alcohol
I saved 950 ml of the extract and, based on the beginning and ending volumes of beer, estimate that I lost approximately 770 ml of the extract down the sink, on the walls, etc. Therefore, the total volume of the extract was 950 + 770, or 1,720 ml.
Again, out of necessity I assumed that the percentage of alcohol in the extract I lost was the same as the percentage of alcohol in the extract I saved. This does not seem to be an outrageous assumption to make. Then I computed the volume of alcohol that was lost with the lost extract:
[770 ml lost extract/1,720 ml total extract volume] x 203.6779104 ml alcohol = 91.1813901199 ml alcohol lost 203.6779104 ml alcohol - 91.1813901199 ml alcohol lost = 112.496520281 ml alcohol in 950 ml saved extract
(extract gravity)(extract volume)=
(SG alcohol)(alcohol volume)+(SG of extract without
alcohol)(alcohol-free extract volume)
(1.039)(950 ml)=(.796)(112.496520281)+(SG)(837.503479719)
987.05=89.5472301436+(SG)837.503479719
897.502769857=(SG)837.503479719
1.07164064578=SG of 950 ml extract without alcohol
The specific gravity of the non-alcohol part of the extract is adjusted for the reduction in volume as follows:
837.503479719 ml (begin volume)/528.949566144 (end volume) x 1.07164064578 (SG non-alcohol part) = 1.11343102248 (SG of non-alcohol part of 600 ml extract)
[Actual SG of extract]/[SG of extract without alcohol] = Efficiency Factor Efficiency Factor x Original Alcohol Vol.=Actual Alcohol Volume 1.069/1.11343102248 = .60829919797 [efficiency factor] .60829919797 x 112.496520281 ml alcohol = 68.4315430613 ml alc.The remaining amount of alcohol in the extract, which is added back to the beer, is 68.4315430613 ml. The total volume of the beer is 7,222.4 ml. Thus, it is simple to do the final calculation and figure out the alcohol content of the beer by volume:
68.4315430613 ml/7,222.4 ml = .0094749035, or approximately 1%The beer therefore still has about 1% alcohol by volume because the extraction method was only about 61% efficient.
Improvement is needed primarily on the efficiency of the heating process to remove alcohol. A longer heating process is necessary. Through a series of other calculations, I estimated that to get a beer of this OG/FG to less than .5% alcohol v/v (the legal definition for a no-alcohol beer), approximately 90% of the extract would have to be evaporated by heating. One-half of the volume that evaporates could be replaced by water without harming the color or flavor of the extract.
I would be very interested in receiving any comments or suggestions anyone might have.
NAF.